Suppose $a,b$ are two natural numbers relatively prime to $n$ and to each other. Assume $n\geq ab+1$. Suppose further that $\frac{a}{b}\equiv k \pmod{n}$ for some $k\in \lbrace 1,2,\dots, n-1\rbrace$ and $\frac{a}{b}\equiv k'\pmod{n+ab}$ for some $k'\in \lbrace 1,2,\dots, n+ab-1\rbrace$.
Question: Is there an elementary proof that the length of the continued fraction of $\frac{n}{k}$ is equal to the length of the continued fraction of $\frac{n+ab}{k'}$?
This came out of a broader result, and for this particular case I can prove it using routine toric geometry, however I would like to know of some elementary tricks to deal with continued fractions.
Here by continued fraction I mean the Hirzebruch continued fraction
$$\frac{n}{k}=a_0-\frac{1}{a_1-\frac{1}{a_2-\cdots}}.$$
For example, when $a=2, b=3$ and $n=17$, we get $k=12$ and $k'=16$, so the fractions are
$$\frac{17}{12}=2-\frac{1}{2-\frac{1}{4-\frac{1}{2}}}\qquad and \qquad\frac{23}{16}=2-\frac{1}{2-\frac{1}{5-\frac{1}{2}}}.$$
Best Answer
Lets call expansions $$\langle x_1,\ldots,x_m\rangle:=\cfrac{1}{x_1-{\atop\ddots\,\displaystyle{-\cfrac{1}{x_m}}}}$$ (as in Perron's book) reduced regular continued fractions (RRCF). Probably they are older then Hirzebruch.
We'll prove more precise statement.
Theorem. If $(n,ab)=1$ and $n>ab$ then RRCF for all numbers $$\left\{\frac{ab^{-1}\pmod{(n+kab)}}{n+kab}\right\}\qquad(k\ge 0)$$ are almost equal: they have equal length and differ only in one partial quotient.
Remark 1. Common factors of $a$ and $b$ can be moved into $k$. If $d=(a,b)$, $a=da_1$, $b=db_1$, then
\begin{gather*} \left\{ \frac{ab^{-1}\pmod{(n+kab)}}{n+kab}\right\} =\left\{ \frac{a_1b_1^{-1}\pmod{(n+kab)}}{n+kab}\right\} \\=\left\{ \frac{a_1b_1^{-1}\pmod{(n+(kd^2)a_1b_1)}}{n+(kd^2)a_1b_1}\right\}. \end{gather*}So we can assume that $(a,b)=1$.Remark 2. The proof will be given in terms of modified continuants $K(x_1,\ldots, x_n)$ (see ``Concrete Mathematics'' for more explanations). These polynomials are defined by initial conditions $$K()=1,\quad K(x_1)=x_1$$ and the following recurrence: $$K(x_1,\ldots, x_n)=x_nK(x_1,\ldots, x_{n-1})-K(x_1,\ldots, x_{n-2})\qquad(n\ge2).$$ (In the usual definition minus must be replaced by plus.) For convenience $K_{-1}:=0$ (empty RRCF is $0$).
In terms of continuants RRCF can be written as $$\langle x_1,\ldots,x_n\rangle=\frac{K(x_2,\ldots, x_n)}{K(x_1,\ldots, x_n)}.$$
Continuant's properties. All these properties can be proved by induction (or from ``Euler’s rule'').
1$^{\circ}.$ $K(x_1,\ldots, x_n)=K(x_n,\ldots, x_{1})$.
2$^{\circ}.$
\begin{gather*} K(x_1,\ldots, x_n,x_{n+1}, \ldots, x_{m+n})\\=K(x_1,\ldots, x_n)K(x_{n+1}, \ldots, x_{m+n})-K(x_1,\ldots, x_{n-1})K(x_{n+2}, \ldots, x_{m+n}) \end{gather*}3$^{\circ}.$ $\begin{vmatrix} K(x_2,\ldots, x_{n-1})&K(x_2,\ldots, x_n) \\ K(x_1,\ldots, x_{n-1})&K(x_1,\ldots, x_n) \end{vmatrix}=-1$. In particular if $$\frac{A}{a}=\left<r_1, \ldots, r_v\right>=\frac{K(r_2, \ldots, r_v)}{K(r_1, \ldots, r_v)}$$ then $$K(r_1, \ldots, r_{v-1})=A^{-1}\pmod{a},\qquad K(r_2, \ldots, r_{v-1})=\frac{AA^{-1}\pmod{a}-1}{a}.$$
4$^{\circ}.$ Euler's identity (see A Short Proof of Euler's Identity for Continuants for additional arguments). (2$^{\circ}$ and 3$^{\circ}$ are special cases of this identity) $$ K(x_1, \ldots, x_{m+n})K(x_{m+1}, \ldots, x_{m+l})-K(x_1, \ldots, x_{m+l})K(x_{m+1}, \ldots, x_{m+n})$$ $$+K(x_1, \ldots, x_{m-1})K(x_{m+l+2}, \ldots, x_{m+n})=0. $$
Proof of the Theorem. For a given $n$ define $a^{-1}:=a^{-1}\pmod{n}$, $b^{-1}:=b^{-1}\pmod{n}$, $0\le a,b\le n-1$ (inverse number is always least possible nonnegative) and $t_a$, $t_b$ such that $aa^{-1}=1+t_an$, $bb^{-1}=1+t_bn$. Let $$ \frac{A}{a}=\left\{\frac{bt_a}{a}\right\}=\left<r_1, \ldots, r_v\right>,\qquad A^{-1}:=A^{-1}\pmod{a};$$ $$\frac{B}{b}=\left\{\frac{at_b}{b}\right\}=\left<q_1, \ldots, q_u\right>,\qquad B^{-1}:=B^{-1}\pmod{b};$$ $$\frac{P(x)}{Q(x)}=\left<q_1, \ldots, q_u,x,r_v, \ldots, r_1\right>. $$ By 2$^{\circ}$, 3$^{\circ}$ and main recurrence $$ Q(x)=K(q_1, \ldots, q_u,x,r_v, \ldots, r_1)=$$ $$xK(q_1, \ldots, q_u)K(r_v, \ldots, r_1)-K(q_1, \ldots, q_{u-1})K(r_v, \ldots, r_1)-K(q_1, \ldots, q_u)K(r_{v-1}, \ldots, r_1)$$ $$=xab-aB^{-1}-bA^{-1}. $$ Hence $$ Q(x)\equiv -bA^{-1}\equiv -b((bt_a)^{-1}\pmod{a})\equiv -t_a^{-1}\equiv n\pmod{a},$$ $$Q(x)\equiv -aB^{-1}\equiv -a((at_b)^{-1}\pmod{b})\equiv -t_b^{-1}\equiv n\pmod{b}.$$ Therefore $$Q(x)\equiv n\pmod{ab},$$ and for some integer $x_0$ we have $Q(x_0)=n$. We know that $n>ab$. It means that $$x_0ab-aB^{-1}-bA^{-1}>ab,$$ so $x_0\ge 2$ and $\left<q_1, \ldots, q_u,x_0,r_v, \ldots, r_1\right>$ is really RRCF. Choosing arbitrary $x=x_0+k$ we'll get progression $n+kab$ as in the statement of the theorem.
Let's check the numerator $P(x)$. Final step $P(x)\equiv ab^{-1}\pmod{Q(x)}$ follows from identity $$bP(x)-BQ(x)=a,$$ which is a special case of Euler's identity. Nevertheless this identity can be verified directly with help of 1$^{\circ}$--3$^{\circ}$: $$ P(x)=K(q_2, \ldots, q_u,x,r_v, \ldots, r_1)=$$ $$xK(q_2, \ldots, q_u)K(r_v, \ldots, r_1)-K(q_2, \ldots, q_{u-1})K(r_v, \ldots, r_1)-K(q_2, \ldots, q_u)K(r_{v-1}, \ldots, r_1)$$ $$=xaB-a\frac{BB^{-1}-1}{b}-BA^{-1},$$ $$ bP(x)-a=B(xab-aB^{-1}-bA^{-1})=BQ(x). $$